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I'm looking for a pointed space $(X,x)$ such that $(X,x)$ is not a good pair, i.e. does not have the homotopy extension property relatively to every space.

Since it seems hard enough to find a space $(X,A)$ without the extension property, I don't know where to look.

I don't even know if or not the Hawaiian earring has this property! Any advice is welcome.

Loulou
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  • It's easier than you think producing a pair $(X, A)$ that does not satisfy the HEP, if for example $X$ is compact but $A$ is not closed in $A$ (because then it is typically not the case that $(I \times A) \cup ({0} \times X)$ would be compact, so that $I \times X$ could not retract onto it). You just want any old pointed space, not necessarily Hausdorff or $T_1$? – user43208 Dec 12 '24 at 16:18
  • Examples I've seen usually are not "algebraic topology-frendly" (discrete image of sequences, Cantor set...). I'm just curious, so I will take any example, but if it could be not too abstract, I would also be glad ;) – Loulou Dec 12 '24 at 16:34

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For $X = \{0, 1, \frac1{2}, \frac1{3}, \ldots\}$, the pair $(X, \{0\})$ does not have the homotopy extension property. If there were a continuous retraction $r: I \times X \to (\{0\} \times X) \cup (I \times \{0\})$ of the inclusion, then $r(t, 1/n) = (0, 1/n)$ for all $n \geq 1$, and $r(\frac1{2}, 0) = (\frac1{2}, 0)$, but clearly $r$ cannot be continuous (by considering the sequence of points $(\frac1{2}, \frac1{n})$).

For $(X, x)$ to satisfy the HEP, I believe it is necessary and sufficient that $X$ be locally contractible at $x$. This would mean that your Hawaiian earring example would also work.

user43208
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